--- title: "Step response of a first-order system" subject: "Instrumentation and Measurement" chapter: 4 key: "semester_4/instrumentation_and_measurement:first-order-step-response-time-constant" source: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/first-order-step-response-time-constant site: BCT Engineering --- # Step response of a first-order system Instrumentation and Measurement · Chapter 4: Dynamic Response of Measurement System - Asked in: year not given, 6 marks - Video: [Unit Step Response of First-Order Systems: Concepts, Derivation, and Graphical Understanding](https://www.youtube.com/watch?v=9_1nprBoEbo) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/first-order-step-response-time-constant - Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_and_measurement/chapter-4 ## Question Derive the response of a first-order measurement system to a step input. Define time constant and show the percentage response at $t = \tau$, $2\tau$, $3\tau$, $4\tau$ and $5\tau$. ## Short answer For a step input qs, the response is qo(t) = K qs (1 - e^(-t/τ)). The time constant τ is the time to reach 63.2% of the final value. At t = τ, 2τ, 3τ, 4τ, 5τ the response is 63.2%, 86.5%, 95.0%, 98.2%, 99.3%. ## Answer The first-order model is $\tau\,\dfrac{dq_o}{dt} + q_o = K q_i$. A step input is $q_i = 0$ for $t<0$ and $q_i = q_s$ (constant) for $t\ge 0$, with $q_o(0) = 0$. #### Derivation The complementary solution is $C e^{-t/\tau}$ and the particular solution is $K q_s$. So $$ q_o(t) = K q_s + C e^{-t/\tau} $$ With $q_o(0) = 0$, $C = -K q_s$, giving $$ q_o(t) = K q_s\left(1 - e^{-t/\tau}\right) $$ The normalized response is $$ \frac{q_o}{K q_s} = 1 - e^{-t/\tau} $$ The error from the final value is $e^{-t/\tau}$, which decays exponentially. #### Time constant The **time constant** $\tau$ is the time taken for the output to reach 63.2% of its final value, $1 - e^{-1}$. Equivalently, it is the time the output would need to reach the final value if it continued at its initial rate. For a thermometer, $\tau = mc/(hA)$. | $t$ | $1 - e^{-t/\tau}$ | Response | |---|---|---| | $\tau$ | 0.632 | 63.2% | | $2\tau$ | 0.865 | 86.5% | | $3\tau$ | 0.950 | 95.0% | | $4\tau$ | 0.982 | 98.2% | | $5\tau$ | 0.993 | 99.3% | ```text q_o/Kqs 1.0 | ____--------- | _-- .632|-----.- | ./ | / | / +------+------+-----> t tau 2tau ``` The settling time to 2% is about $4\tau$ and to 5% about $3\tau$. A small $\tau$ means a fast instrument. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The step response of a first-order system is A. K qs e^(-t/tau) B. K qs (1 - e^(-t/tau)) C. K qs (1 - e^(-t)) D. K qs sin(t/tau) Answer: B. Solving with qo(0) = 0 gives qo = K qs (1 - e^(-t/tau)). (Key point: Step response) 2. The time constant tau is the time for the step response to reach A. 50% of final value B. 63.2% of final value C. 95% of final value D. 100% of final value Answer: B. At t = tau, 1 - e^-1 = 0.632. (Key point: Time constant meaning) 3. What is the response at t = 3tau? A. 86.5% B. 95.0% C. 98.2% D. 99.3% Answer: B. 1 - e^-3 = 0.950. (Key point: Percent at 3 tau) 4. What does a small time constant imply? A. A slow instrument B. A fast instrument C. Larger overshoot D. Higher static sensitivity Answer: B. Small tau means the output approaches the final value quickly. 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